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Consider two processors P1 and P2 executing the same instruction set. Assume that under identical conditions, for the same input, a program running on P2 takes 25% less time but incurs 20% more CPI (clock cycles per instruction) as compared to the program running on P1. If the clock frequency of P1 is 1GHz, then the clock frequency of P2 (in GHz) is _________.
  • a)
    1.6
  • b)
    3.2
  • c)
    1.2
  • d)
    0.8
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Consider two processors P1 and P2 executing the same instruction set. ...
For P1 clock period = 1ns Let clock period for P2 be t.
Now consider following equation based on specification 7.5 ns = 12*t ns
We get t and inverse of t will be 1.6GHz
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Most Upvoted Answer
Consider two processors P1 and P2 executing the same instruction set. ...
Given:
- Two processors P1 and P2 executing the same instruction set.
- For the same input, a program running on P2 takes 25% less time but incurs 20% more CPI as compared to the program running on P1.
- Clock frequency of P1 is 1GHz.

To find: The clock frequency of P2.

Assumption: The number of instructions executed by both processors is the same.

Solution:
Let's assume that P1 takes T1 time to execute the program and P2 takes T2 time to execute the same program.

According to the question, T2 = 0.75T1 (as P2 takes 25% less time than P1)

Also, CPI for P2 is 20% more than CPI for P1. Let CPI for P1 be x. Then, CPI for P2 will be 1.2x.

Now, we know that the time taken by a processor to execute a program is given by:

Time taken = Number of instructions x CPI / Clock frequency

As the number of instructions executed by both processors is the same, we can equate the above equation for both processors.

T1 = Number of instructions x x / 1GHz (for P1)
T2 = Number of instructions x 1.2x / f (for P2)

where f is the clock frequency of P2.

Now, we can equate T1 and T2 and solve for f.

T1 = T2
Number of instructions x x / 1GHz = Number of instructions x 1.2x / f
f = 1GHz / 1.2
f = 0.83 GHz

Therefore, the clock frequency of P2 is 0.83 GHz or 830 MHz (approx).

Answer: Option A (1.6 GHz) is incorrect. The correct answer is Option D (0.8 GHz).
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